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4vir4ik [10]
2 years ago
8

What are the zeros of the function? f(x)=+-6x

Mathematics
1 answer:
Step2247 [10]2 years ago
7 0
The function given is a line with one zero at the center (0,0)
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Why is the vertex of vertex form y=a(x-h)^2+k (h,k) rather than (-h,k)? If that was y=2(x-5)^2+6, the vertex would be at (5,6) w
Sindrei [870]

9514 1404 393

Answer:

  (5, 6) is (h, k)

Step-by-step explanation:

Vertex form is an instance of the transformation of parent function f(x) = x². It is vertically scaled by a factor of 'a', and translated so the vertex is point (h, k). That is, the transformed vertex is h units right and k units up from that of the parent function (0, 0).

Parent:

  f(x) = x^2

Transformed:

  f(x) = a(x -h)^2 +k

__

When you compare the form to your specific instance, you need to pay attention to what it is that you're comparing. As the attachment shows, ...

  • a = 2
  • -h = -5   ⇒   h = 5
  • k = 6

Hence the vertex is (h, k) = (5, 6). The second attachment shows this on a graph.

7 0
3 years ago
In a triangle what is always opposite the angle with the greatest measure
Llana [10]
The longest side

Think about it. The widest angle would leave a resulting long side to reach the ends of the angle 

If you look at the picture, the side across from the widest angle has the longest length

5 0
2 years ago
Read 2 more answers
I need help with this please
Harlamova29_29 [7]
-12y+14-9y=14
-21y=0
y=0, x=7
3 0
3 years ago
Two professional baseball teams played a four game series. Attendance for the first three games was 126,503 people, what was the
ioda

Answer: 44,815 people; 126,503 + n = 171,318; n = 171,318 − 126,503

6 0
3 years ago
The equation r(t)= (2t)i + (2t-16t^2)j is the position of a particle in space at time t. Find the angle between the velocity and
ankoles [38]

Answer:

The answer is 135 degrees.

Step-by-step explanation:

As we are given the position. If we take the <u>derivative</u>, we get the velocity vector. If we take the <u>derivative</u> again, we find the acceleration vector of the particle.

r(t)=(2t)i+(2t-16t^{2})j

V(t)=2i+(2-32t)j

a(t)=-32j

At time t=0;

v(t)=2i+2j

a(t)=-32j

As i attach in the picture the angle between the velocity and acceleration vector is (45+90)=135 degrees

4 0
3 years ago
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